The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $23.8$ years; the standard deviation is $4.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $36.1$ years.
Solution: $23.8$ $19.7$ $27.9$ $15.6$ $32$ $11.5$ $36.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $23.8$ years. We know the standard deviation is $4.1$ years, so one standard deviation below the mean is $19.7$ years and one standard deviation above the mean is $27.9$ years. Two standard deviations below the mean is $15.6$ years and two standard deviations above the mean is $32$ years. Three standard deviations below the mean is $11.5$ years and three standard deviations above the mean is $36.1$ years. We are interested in the probability of a tiger living less than $36.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $11.5$ years and the other half $({0.15\%})$ will live longer than $36.1$ years. The probability of a particular tiger living less than $36.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.